Curl and partial derivative identity ($\frac{d}{dt} = \frac{\partial }{\partial t} + (\vec{v} \cdot \vec{\nabla})$)

Question

On this site I found the following identity (labeled $13$ there) :

$$\frac{d}{dt} = \frac{\partial }{\partial t} + (\vec{v} \cdot \vec{\nabla})$$

Where $\vec{v}$ denotes the velocity vector. Does this make sense ? How does one derive this result?

Answer 1

Instead of deriving it, I will answer the question about whether it makes sense from a dimensional perspective, as you might see in Street-Fighting Mathematics from the MIT Press.

The $d/dt$ operator has dimensions of $\text{time}^{-1}$. Each component of the velocity vector has dimensions of $\text{length}\cdot\text{time}^{-1}$. Look at the $\vec \nabla$ operator. What are its dimensions? Unpacking the partials, we see that each partial is taken with respect to the spatial variables, and $\partial/\partial (x,y,z)$ has dimensions of $\text{length}^{-1}$. The dot product means we multiply $\vec v$ and $\vec \nabla$ component-wise and sum.

From a dimensional perspective: $$ \frac{d}{dt} = \frac{\partial }{\partial t} + (\vec{v} \cdot \vec{\nabla}) \\ \text{time}^{-1} = \text{time}^{-1} + (\text{length}\cdot\text{time}^{-1} )\cdot\text{length}^{-1} = \text{time}^{-1} + \text{time}^{-1} = \text{time}^{-1}.\\ $$ In this sense the operator identity makes sense.

If we want a more physical description of why the equality holds, consider that the operator $d/dt$ is a measure of the "net-change" in our physical quantity. Our physical quantity is itself a function of spatial and temporal variables. The $\partial/\partial t$ operator catches, in one go, the dependence on our quantity with respect to its temporal part. The quantity also has a spatial aspect to it. With dimensional analysis, we can see how the product of $\vec v$, a test-particle's velocity and $\vec \nabla$ captures the rate of change of the quantity with respect to time in terms of its spatial variables.

Note that it makes sense that the right-hand side does not admit any multiplicative factors because the two operators on the right-hand side together make up the entirety of the change of our physical quantity with respect to its pure temporal part, as well as changes in time with respect to changes in its spatial part.

Even more physical intuition: we expect that changes of our quantity in time should only come out of the $\vec v\cdot \vec \nabla$ term if there is motion so $\vec v \ne \vec 0$. If the motion is slow, so $v \approx 0$, where $v$ is the speed of the quantity, the contribution from $\vec v\cdot \vec \nabla$ should be small, and if $v \gg 0$ and the quantity does change with variations in its spatial variables, the contribution from $\vec v\cdot \vec \nabla$ should be large, since the test-particle is moving quickly through the quantity along one of its flow lines.

Likewise, if the quantity does not change on its own in time, that is, it has no purely temporal aspect, then we expect no contribution from $\partial/\partial t$, since this operator captures only explicit time-dependence unlike $\vec v\cdot \vec \nabla$, which captures an implicit dependence on time that is a virtue of the test-particle's motion through the quantity (be it a field, or a fluid, or what have you).

In the ways I have described, the velocity $\vec v$ of the test-particle works to sniff out information in the way the quantity changes in time, an implicit characteristic of the quantity's time dependence.

Answer 2

Here we are looking at some kind of fluid flow and the variable, say $\phi$, so this says that $\frac{d\phi}{dt}= \frac{\partial \phi}{\partial t}+ (\vec{v}\cdot\vec{\nabla}\phi)$, is a property of the fluid (density, temperature, etc.). $\frac{\partial\phi}{\partial t}$ is the derivative of that property of the fluid at a particular point- you hold a thermometer in the fluid at a fixed (x, y) point in the flow. $\frac{d\phi}{dt}= \frac{\partial \phi}{\partial t}+ (\vec{v}\cdot\vec{\nabla}\phi)$ is the rate of change of that property as you move along with the fluid- your thermometer is attached to small "boat" that flows with the fluid.

Answer 3

The equality follows immediately from the chain rule.

Let $\phi(x,y,z,t)$ be a differentiable function of $x,y,z,$ and $t$. Further, suppose that $x=x(t)$, $y=y(t)$, and $z=z(t)$ are differentiable functions of $t$.

Denoting $\Phi(t)=\phi(x(t),y(t),z(t),t)$, the derivative, $\Phi'(t)$, of $\Phi(t)$ is given by the chain rule as

$$\begin{align}\Phi'(t)&=\frac{d\Phi(t)}{dt}\\\\ &=\left.\left(\frac{\partial \phi(x,y,z,t)}{\partial t}\right)\right|_{(x,y,z)=(x(t),y(t),z(t))}\\\\ &+\left.\left(\frac{\partial \phi(x,y,z,t)}{\partial x}\frac{dx(t)}{dt}\right)\right|_{(x,y,z)=(x(t),y(t),z(t))}\\\\ &+\left.\left(\frac{\partial \phi(x,y,z,t)}{\partial y}\frac{dy(t)}{dt}\right)\right|_{(x,y,z)=(x(t),y(t),z(t))}\\\\ &+\left.\left(\frac{\partial \phi(x,y,z,t)}{\partial z}\frac{dz(t)}{dt}\right)\right|_{(x,y,z)=(x(t),y(t),z(t))}\\\\ &=\left.\left(\nabla \phi(x,y,z,t)\cdot \vec v(t)\right)\right|_{(x,y,z)=(x(t),y(t),z(t))} \end{align}$$

where $\vec v(t)=\hat x\frac{dx(t)}{dt}+\hat y\frac{dy(t)}{dt}+\hat z\frac{dz(t)}{dt}$.

Hence, the operator $\frac{d}{dt}$ is equivalent to the operator $\frac{\partial}{\partial t}+\vec v \cdot \nabla$.